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MATH 355 MIDTERM SOLUTIONS
Here are the solutions to your recent midterm. Enjoy!
(1) (a) and (b) were given in class and/or in the solutions to the homework and/or the
book.
(c) We use double-inclusion. Let (s, t) ∈ S × T . Then (s, t) ∈ Ts for this s, and
so (s, t) ∈ ∪x∈S Tx . Now let y ∈ ∪x∈S Tx . Then there exists an x ∈ S for which
y ∈ Tx . So, y = (x, t) for some t ∈ T , and so y = (x, t) ∈ S × T .
(d) We use double-inclusion. Clearly {2x|x ∈ R} ⊆ R. Now let y ∈ R. Let
x = y2 , so that y = 2 y2 = 2x ∈ {2x|x ∈ R}.
(2) (a) and (b) were done in class, and a solution to (c) appears in our homework
solutions.
(d) (i) f (A) = {4, 16, 36, 64} (ii) f −1 (B) = {1, 2}, (iii) Yes, since there are only
two square roots to any number, a positive one and a negative one (unless the
number is 0, which is neither in the domain or range of this function), and the
domain only contains positive numbers, there is only one element in the domain
that maps to each element of the range.
(3) (a) Clearly, if |X| = 1, then X contains only one element, and this element must
be the largest in the set. Now suppose that any set with cardinality n contains a
largest element. Let X be a set with |X| = n + 1. Choose any x ∈ X, and consider
the set X − {x}. |X − {x}| = n, and so there is a largest element y ∈ X − {x}. The
set X is the disjoint union of X − {x} and {x}. Let z be the larger number between
x and y. Then z is larger than every element of X since z ≥ y and y is larger than
every element of X − {x}, and z ≥ x by construction. So z is the largest element
of X.
(b) This was proven in class.
(c) This was proven in class, although a solution could be put here: Using 1(c)
above, S × T is a union of sets. This union is indexed by S, so the union is a
countable one. Each set Tx ∼ T via the bijection (x, t) 7→ t. Since T is countable,
so must Tx be countable, and so each set being unioned is countable. Thus, S × T
is the countable union of countable sets, and is therefore countable by one of our
results.
(d) This was proven in class.
(e) Suppose the irrationals were countable. Then the irrationals unioned with Q
would also be countable, since Q is countable as well. But this union is all of R,
which is known to be uncountable. This is a contradication.
(f) We prove this by induction. Clearly, if we cross product only one copy of
N, this is a countable set. Now suppose that the cross product of n copies of N is
countable. The cross product of n + 1 copies of N is the cross product of N with the
1
2
MATH 355 MIDTERM SOLUTIONS
cross product of n copies of N, and both of these are known to be countable: the
first since N is countable, and the second by the induction hypothesis. According
to 3(c), such a cross product is again countable, proving the claim. (To answer the
extra credit problem, such a countably infinite cross product of N’s is uncountable:
it contains the countably infinite cross product of {1, . . . , 9, 10}, and there is an onto
map from this cross product onto [0, 1] by expressing each sequence of numbers (minus one) as the decimal expansion of a number in [0, 1]. Since [0, 1] is uncountable,
this subset is uncountable, hence the original cross product is uncountable.)
(g) Set
x
if x 6= 2−n+1 for n ∈ N
f (x) =
.
−n+2
2
if x = 2−n+1
This is clearly 1-1, since the only thing this function changes is an advance of the
sequence 2−n+1 (similar to a proof of ours), and is onto, since every one of the
outputs is in (0, 1): if x is not one of 2−n+1 , then x is mapped from itself. If x is
one of the 2−n+1 , then it is mapped to by the previous one in the sequence. The
only exception to this is that when n = 1, this does not have a predecessor to be
mapped from, and this is the output 2−1+1 = 20 = 1, and 1 ∈
/ (0, 1).
π
(h) The function f (x) = π2 x is a bijection from (−1, 1) onto ( −π
2 , 2 ). The tanπ
gent function g(x) = tan x is a bijection from ( −π
2 , 2 ) onto R. Therefore their
composition g ◦ f is a bijection from (−1, 1) onto R.